Problem: What is the value of the following logarithm? $\log_{8} 8$
If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $8^{y} = 8$ Any number raised to the power $1$ is simply itself, so $8^{1} = 8$ and thus $\log_{8} 8 = 1$.